Question: $f(x)=x^2+3x+2$ What is the average rate of change of $f$ over the interval $[-1,-1+h]$, in terms of $h$, where $h\neq 0$ ? Your answer must be fully expanded and simplified.
Answer: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We can calculate that $f(-1)=0$. We are interested in the average rate of change of $f(x)=x^2+3x+2$ over the interval $[-1,-1+h]$ : $\begin{aligned} &\phantom{=}\dfrac{f(-1+h)-f(-1)}{(-1+h)-(-1)} \\\\ &=\dfrac{(-1+h)^2+3(-1+h)+2-0}{-1+h-(-1)} \\\\ &=\dfrac{(-1+h)^2+3(-1+h)+2}{h} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} &\phantom{=}\dfrac{(-1+h)^2+3(-1+h)+2}{h} \\\\ &=\dfrac{1-2h+h^2-3+3h+2}{h} \\\\ &=\dfrac{h^2+h}{h} \\\\ &=\dfrac{h(h+1)}{h} \\\\ &=h+1\text{, for }h\neq 0 \end{aligned}$ Since we are given that $h\neq 0$, the average rate of change of the function is $h+1$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.